Question

我有三个表：Schools（id，school_name），Schools_Focuses（id，school_id，focus_id）和Focuses（id，focus），并希望在我的Schools模型上创建一个返回所有相关焦点的方法。

我可以执行我想要对此SQL QUERY执行的操作：

SELECT focus FROM focuses INNER JOIN schools_focuses ON focuses.id = schools_focuses.focus_id INNER JOIN schools ON schools.id = schools_focuses.school_id WHERE schools.id = 36;

型号代码：

// Define Models
exports.School = School = Bookshelf.PG.Model.extend({
  tableName: 'schools',

  focuses: function() {
    return this.hasMany(Focus).through(Schools_Focuses);
  }
});

错误：

可能未处理错误：列focus.schools_focuse_id不存在，sql：选择“focused”。*，“schools_focuses”。“id”为“_pivot_id”，“schools_focuses”。“school_id”为“_pivot_school_id”来自“focus” “inner join”schools_focuses“on”schools_focuses“。”id“=”focus“。”schools_focuse_id“where”schools_focuses“。”school_id“in（？）

我不希望在Focuses中有这个额外的列（focused.schools_focuse_id），因为焦点可以归属于多个学校。

如何正确设置？我玩过外键和其他键用于hasMany（）和through（），但没有运气。

谢谢！

Answer 1

听起来你需要使用belongsToMany()而不是hasMany（）。

// Define Models
exports.School = School = Bookshelf.PG.Model.extend({
  tableName: 'schools',

  focuses: function() {
    return this.belongsToMany(Focus, 'schools_focuses');
  }
});

第二个参数是必需的，因为连接表不是alpha顺序（'focused_schools'）。如果在识别连接键（school_id，focus_id）时遇到问题，您也可以将这些键作为参数覆盖。此外，由于它在内部使用连接表，因此您无需为Schools_Focuses创建单独的模型。

Answer 2

如果我们不需要为Schools_Focuses创建单独的模型，那么我们如何从联结表中查询数据，即：

'SELECT * FROM focuses_schools WHERE focus_id = 2 & school_id = 1'

建立“多对多”关系

2 个答案: